An elastic ball of mass m falls from a height h on an Aluminium disc of area A floating in a mercury pool. If the collision is perfectly elastic, the momentum transferred to the disc is :

Options

A. `sqrt(2mgh)`
B. `2sqrt(mgh)`
C. `m sqrt(2gh)`
D. `2m sqrt(2gh)`

Answer

To solve the problem of momentum transferred to the disc when an elastic ball of mass \( m \) falls from a height \( h \) and collides with an aluminum disc, we can follow these steps:

### Step 1: Determine the velocity of the ball just before the collision
When the ball falls from a height \( h \), we can use the equation of motion to find its velocity just before it hits the disc. The velocity \( v \) just before impact can be calculated using the formula:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Understand the nature of the collision
Since the collision is perfectly elastic, both momentum and kinetic energy are conserved. In an elastic collision, the coefficient of restitution \( e \) is equal to 1.

### Step 3: Apply conservation of momentum
Let:
- \( u_1 = \sqrt{2gh} \) (initial velocity of the ball)
- \( u_2 = 0 \) (initial velocity of the disc, as it is floating)

Using the conservation of momentum for the elastic collision, we can express the final velocities after the collision:
\[ m u_1 + M u_2 = m v_1 + M v_2 \]
where \( v_1 \) and \( v_2 \) are the final velocities of the ball and the disc, respectively. Since the disc is floating, we can assume its mass \( M \) is much larger than \( m \), and thus the velocity of the disc after the collision will be very small.

### Step 4: Use the coefficient of restitution
The coefficient of restitution for a perfectly elastic collision is given by:
\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]
Substituting \( e = 1 \):
\[ 1 = \frac{v_2 - v_1}{u_1 - 0} \]
This implies:
\[ v_2 - v_1 = u_1 \]
Thus, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v_1 + u_1 \]

### Step 5: Calculate the change in momentum
The change in momentum \( \Delta p \) transferred to the disc can be calculated as:
\[ \Delta p = m(v_1 - u_1) + M(v_2 - u_2) \]
Since \( u_2 = 0 \) and \( v_1 \) will be negative after the collision (the ball rebounds), we can substitute:
\[ \Delta p = m(-\sqrt{2gh} - \sqrt{2gh}) + Mv_2 \]
This simplifies to:
\[ \Delta p = 2m\sqrt{2gh} \]

### Final Answer
Thus, the momentum transferred to the disc is:
\[ \Delta p = 2m\sqrt{2gh} \]